{"id":19924,"date":"2017-02-03T23:33:28","date_gmt":"2017-02-03T23:33:28","guid":{"rendered":"http:\/\/www.oracletutoring.ca\/blog\/?p=19924"},"modified":"2017-02-04T04:17:33","modified_gmt":"2017-02-04T04:17:33","slug":"ac-electricity-electronics-series-high-pass-filter","status":"publish","type":"post","link":"https:\/\/www.oracletutoring.ca\/blog\/ac-electricity-electronics-series-high-pass-filter\/","title":{"rendered":"AC Electricity, electronics:  series high-pass filter"},"content":{"rendered":"<h1>The tutor gives background along with a basic explanation of how a high-pass filter works.<\/h1>\n<p>A high-pass filter will send along high frequency signals but block low frequencies.  It can do so because the impedance, X<sub>c<\/sub>, of a capacitor of capacitance C, at frequency f, is<\/p>\n<p style=\"text-align:center\">X<sub>c<\/sub>=1\/(2\u03c0fC)<\/p>\n<p>At very high frequency, therefore, X<sub>c<\/sub>\u22480<\/p>\n<p>The total impedance, Z, of a resistor R and capacitor C in series is given by<\/p>\n<p>Z=(R<sup>2<\/sup> + X<sub>c<\/sub><sup>2<\/sup>)<sup>0.5<\/sup><\/p>\n<p>Therefore, with input voltage V, and instantaneous current i, Ohm&#8217;s Law gives<\/p>\n<p>V=i(R<sup>2<\/sup> + X<sub>c<\/sub><sup>2<\/sup>)<sup>0.5<\/sup><\/p>\n<p>The voltage across the resitor is<\/p>\n<p>V<sub>R<\/sub>=iR<\/p>\n<p>Therefore, the ratio of the voltage across the resistor to the source is<\/p>\n<p>V<sub>R<\/sub>\/V = iR\/i(R<sup>2<\/sup> + X<sub>c<\/sub><sup>2<\/sup>)<sup>0.5<\/sup> = R\/(R<sup>2<\/sup> + X<sub>c<\/sub><sup>2<\/sup>)<sup>0.5<\/sup><\/p>\n<p>As mentioned earlier, for high frequency, X<sub>c<\/sub>\u22480, giving<\/p>\n<p>V<sub>R<\/sub>\/V \u2248 R\/(R<sup>2<\/sup>)<sup>0.5<\/sup> = 1<\/p>\n<p>With a high-pass filter, the maximum output voltage for a high frequency signal equals the input voltage.<\/p>\n<p>Convention says that the critical frequency, f<sub>c<\/sub>, is that at which V<sub>R<\/sub>\/V = 1\/(2)<sup>0.5<\/sup> = 0.707, which occurs when X<sub>c<\/sub> = R:<\/p>\n<p>V<sub>R<\/sub>\/V = R\/(R<sup>2<\/sup> + X<sub>c<\/sub><sup>2<\/sup>)<sup>0.5<\/sup> = R\/(R<sup>2<\/sup> + R<sup>2<\/sup>)<sup>0.5<\/sup> = R\/(2R<sup>2<\/sup>)<sup>0.5<\/sup><\/p>\n<p>which leads to<\/p>\n<p>V<sub>R<\/sub>\/V = 1\/2<sup>0.5<\/sup> = 0.707<\/p>\n<p>Therefore, a high-pass filter will pass any frequency higher than the critical frequency f<sub>c<\/sub>, where f<sub>c<\/sub> is calculated from<\/p>\n<p>X<sub>c<\/sub> = R<\/p>\n<p>1\/(2\u03c0f<sub>c<\/sub>C) = R<\/p>\n<p>1 = 2\u03c0f<sub>c<\/sub>CR<\/p>\n<p>1\/(2\u03c0CR) = f<sub>c<\/sub><\/p>\n<p>By that reasoning, a 10k\u2126 resistor, in series with a 12pF (picoFarad) capacitor, placed in series, should produce a high-pass filter with critical frequency f<sub>c<\/sub> = 1.3MHz.  The output would be read across the resistor R.<\/p>\n<p>Source:<\/p>\n<p><a href=\"http:\/\/www.electronics-tutorials.ws\/filter\/filter_3.html\">www.electronics-tutorials.ws<\/a><\/p>\n<p><a href=\"https:\/\/ecee.colorado.edu\/~mcclurel\/resistorsandcaps.pdf\">ecee.colorado.edu<\/a><\/p>\n<p>Serway, Raymond A.  <u>Physics for Scientists and Engineers<\/u> with modern physics.  Toronto: Saunders College Publishing, 1986.<\/p>\n<p>Jack of <a href=\"https:\/\/www.oracletutoring.ca\">Oracle Tutoring by Jack and Diane,<\/a> Campbell River, BC.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The tutor gives background along with a basic explanation of how a high-pass filter works. A high-pass filter will send along high frequency signals but block low frequencies. It can do so because the impedance, Xc, of a capacitor of &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"more-link\" href=\"https:\/\/www.oracletutoring.ca\/blog\/ac-electricity-electronics-series-high-pass-filter\/\"> <span class=\"screen-reader-text\">AC Electricity, electronics:  series high-pass filter<\/span> Read More &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1023,297,1156],"tags":[],"class_list":["post-19924","post","type-post","status-publish","format-standard","hentry","category-electricity","category-electronics","category-engineering"],"_links":{"self":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/19924","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/comments?post=19924"}],"version-history":[{"count":38,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/19924\/revisions"}],"predecessor-version":[{"id":19962,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/19924\/revisions\/19962"}],"wp:attachment":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/media?parent=19924"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/categories?post=19924"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/tags?post=19924"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}